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In quad. ADCF ,
AC = DF , AC || DF ( given)
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
therefore, ADCF is a ||gm
thus, AD = CF , AD || CF (opposite side of ||gm is equal and parallel to each other) ____(i)
In quad. ADEB ,
AB = DE , AB || DE (given)
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
Therefore, ADEB is a ||gm.
Thus, AD = EB , AD || EB ( opposite side of ||gm )
From (i) and (ii)...
AD = EB = CF , AD || CF || EB _____(iii)
Now,
In quad. BCFE ,
CF = BE , CF || BE ( proved in eq.{iii})
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
Therefore , quad. BCFE is a ||gm
Thus , BC = EF , BC || EF ( opposite side of ||gm)
Hence, Proved
AC = DF , AC || DF ( given)
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
therefore, ADCF is a ||gm
thus, AD = CF , AD || CF (opposite side of ||gm is equal and parallel to each other) ____(i)
In quad. ADEB ,
AB = DE , AB || DE (given)
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
Therefore, ADEB is a ||gm.
Thus, AD = EB , AD || EB ( opposite side of ||gm )
From (i) and (ii)...
AD = EB = CF , AD || CF || EB _____(iii)
Now,
In quad. BCFE ,
CF = BE , CF || BE ( proved in eq.{iii})
(since : In quad. if one pair of opposite side is parallel and equal to each other then quad. is a ||gm)
Therefore , quad. BCFE is a ||gm
Thus , BC = EF , BC || EF ( opposite side of ||gm)
Hence, Proved
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